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anonymous

  • 4 years ago

Rationalize the denominator: cubic root of 2y^4/6x^4

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  1. anonymous
    • 4 years ago
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    \[\sqrt[3]{2y ^{4}/6x ^{4}}\]

  2. anonymous
    • 4 years ago
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    My instinct is to go down instead of making it \[x ^{64}\] because that would look dumb, unless it's correct but I've never had to decrease... do I just multiply by a negative exponent?

  3. anonymous
    • 4 years ago
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    =cube root (2y^4)/cube root(6x^4) =cube root(2y^4)*cuberoot(6x^4)/cube root(6x^4)*cube root(6x^4) =cube root(2y^4)*cuberoot(6x^4)/6x^4 basically separate top and bottom roots multiply by conjugates fix ugly maths

  4. jhonyy9
    • 4 years ago
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    cuberoot(y/3x)^4 =(y/3x)cuberoot(y/3x)=(ycuberoot3xy)/3x

  5. jhonyy9
    • 4 years ago
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    2/6 simplifie by 2 and get 1/3

  6. jhonyy9
    • 4 years ago
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    ok ?

  7. anonymous
    • 4 years ago
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    Hmm, well that is handy, didn't know oyu could simplify it but what would the correct conjugate be?

  8. anonymous
    • 4 years ago
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    you*

  9. jhonyy9
    • 4 years ago
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    good luck bye

  10. jhonyy9
    • 4 years ago
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    1/sqrt3 =(sqrt3)/3 ---like example

  11. jhonyy9
    • 4 years ago
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    ok ?

  12. anonymous
    • 4 years ago
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    No actually, I'm confused... can you just tell me what I need to multiply by to get rid of the radical in the denominator?

  13. jhonyy9
    • 4 years ago
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    1/sqrt3 so when there is radical in denominator multiplie numerator and denominator by conjugate so in this case by sqrt3 and in this way you get in denominator 3 and in numerator sqrt3

  14. jhonyy9
    • 4 years ago
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    because sqrt3 * sqrt3 =3

  15. anonymous
    • 4 years ago
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    I know... but what would the conjugate be. I know how to do all this already i get it I just need to know if the conjugate I am using is correct...

  16. jhonyy9
    • 4 years ago
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    your answer not is right

  17. jhonyy9
    • 4 years ago
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    check my answer step by step

  18. razor99
    • 4 years ago
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    we have to simplify that.

  19. anonymous
    • 4 years ago
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    Ok look, the denominator is cubic root of 6x^4 .... so what do I multiply it by to take a perfect cubic root?

  20. anonymous
    • 4 years ago
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    sorry that is wrong

  21. anonymous
    • 4 years ago
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    \[\sqrt[3]{36x^2}\]

  22. saifoo.khan
    • 4 years ago
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    If sat is here, then there was no need of calling me! :)

  23. anonymous
    • 4 years ago
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    then you will get in the denominator a perfect cube, namely \[\sqrt[3]{216x^6}\]

  24. anonymous
    • 4 years ago
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    He just got here thanks anyway

  25. anonymous
    • 4 years ago
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    you got this?

  26. anonymous
    • 4 years ago
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    you have \[\sqrt[3]{6x^4}\] so to make it a perfect cube, you need the exponent on the x to be 6, and you have to multiply by 6^2=36 to make the constnant a cube

  27. anonymous
    • 4 years ago
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    Wait why does the exponent for x have to be 6?

  28. anonymous
    • 4 years ago
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    because you have a cube root so you need a number divisible by 3

  29. anonymous
    • 4 years ago
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    i mean an exponent divisible by 3

  30. anonymous
    • 4 years ago
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    \[\sqrt[3]{216x^6}=6x^2\]

  31. anonymous
    • 4 years ago
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    So that would come out as \[6x ^{2}\] ?

  32. anonymous
    • 4 years ago
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    yes

  33. anonymous
    • 4 years ago
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    Ah thank you that is ALL I needed, I guess nobody understood what I meant. Thank you very much

  34. anonymous
    • 4 years ago
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    yw

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